Ask question

Ask question

All categories

3 years ago

14

The tea chest needs to send the sales tax that it has collected to the state. The total sales, including 4% sales tax are $7115.

68. How much in sales tax does the store need to send to the state?

1 answer:

Elina [12.6K]3 years ago

8 0

.04*7115.68=284.6272

$284.62 or $284.63 (depending on how lenient your teacher is with rounding)

You might be interested in

What is the value of x?

Angelina_Jolie [31]

Answer:

x = 5.4

Step-by-step explanation:

3/x = 4/7.2

cross-multiply:

4x = 21.6

x = 5.4

5 0

3 years ago

Plz help me find zero x on the triangle and show work thanks

netineya [11]

Answer:

x= 30 degrees

Step-by-step explanation:

This is an isosceles triangle as indicates by the lines on the sides.

Since the sides lengths are equal, the base angles are equal

x= 30 degrees

6 0

3 years ago

Find the solutions of the quadratic equation 3x^2 + 9x + 17 = 0

kirill115 [55]

Answer:

x=

6

9−

285

=

2

3

−

6

1

285

=−1.314

x=

6

9+

285

=

2

3

+

6

1

285

=4.314

Step-by-step explanation:

8 0

3 years ago

Is it a function and what is the Domain and the Range

Schach [20]

Answer:

a-is a function

the domain is:[-3,1]

the range is :[-4,0]

b- is not a function

7 0

3 years ago

Read 2 more answers

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

3 years ago